Expectation of F-Distribution

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n, m$ be strictly positive integers.

Let $X \sim F_{n, m}$ where $F_{n, m}$ is the F-distribution with $\tuple {n, m}$ degrees of freedom.


Then the expectation of $X$ is given by:

$\expect X = \dfrac m {m - 2}$

for $m > 2$, and does not exist otherwise.


Proof

Let $Y$ and $Z$ be independent random variables.

Let $Y \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.

Let $Z \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom.

Then, from Quotient of Independent Random Variables with $\chi^2$ Distribution Divided by Degrees of Freedom has $F$-Distribution:

$\dfrac {Y / n} {Z / m} \sim F_{n, m}$

So we aim to compute:

$\expect {\dfrac {Y / n} {Z / m} }$

Let $f_Y$ and $f_Z$ be the probability density functions of $Y$ and $Z$ respectively.

Let $f_{Y, Z}$ be the joint probability density function of $Y$ and $Z$.

From Condition for Independence from Joint Probability Density Function, we have for each $y, z \in \R_{\ge 0}$:

$\map {f_{Y, Z} } {y, z} = \map {f_Y} y \map {f_Z} z$

We therefore have:

\(\ds \expect {\dfrac {Y / n} {Z / m} }\) \(=\) \(\ds \int_0^\infty \int_0^\infty \frac {y / n} {z / m} \map {f_{Y, Z} } {y, z} \rd y \rd z\)
\(\ds \) \(=\) \(\ds \frac m n \int_0^\infty \int_0^\infty \frac y z \map {f_Y} y \map {f_Z} z \rd y \rd z\)
\(\ds \) \(=\) \(\ds \frac m n \paren {\int_0^\infty \frac {\map {f_Z} z} z \rd z} \paren {\int_0^\infty y \map {f_Y} y \rd y}\) rewriting
\(\ds \) \(=\) \(\ds \frac m n \paren {\frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z} \paren {\frac 1 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty y^{n / 2} e^{-y / 2} \rd y}\) Definition of Chi-Squared Distribution

Note that the integral converges:

$\displaystyle \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z$

if and only if:

$\dfrac m 2 - 2 > -1$

That is:

$m > 2$

With that, we have:

\(\ds \frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z\) \(=\) \(\ds \frac 1 {2^{m / 2 - 1} \map \Gamma {\frac m 2} } \int_0^\infty \paren {2 u}^{m / 2 - 2} e^{-u} \rd u\) substituting $z = 2 u$
\(\ds \) \(=\) \(\ds \frac {2^{m / 2 - 2} } {2^{m / 2 - 1} \map \Gamma {\frac m 2} } \int_0^\infty u^{m / 2 - 2} e^{-u} \rd u\)
\(\ds \) \(=\) \(\ds \frac 1 2 \frac {\map \Gamma {\frac m 2 - 1} } {\map \Gamma {\frac m 2} }\) Definition of Gamma Function
\(\ds \) \(=\) \(\ds \frac {\map \Gamma {\frac m 2 - 1} } {\paren {\frac m 2 - 1} \map \Gamma {\frac m 2 - 1} }\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds \frac 1 {m - 2}\)

The integral:

$\displaystyle \int_0^\infty y^{n / 2} e^{-y / 2} \rd y$

converges if and only if $n > -2$.

This is ensured by the fact that $n \in \N$.

With that, we have:

\(\ds \frac 1 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty y^{n / 2} e^{-y / 2} \rd y\) \(=\) \(\ds \frac 2 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty \paren {2 v}^{n / 2} e^{-v} \rd v\) substituting $y = 2 v$
\(\ds \) \(=\) \(\ds \frac {2^{n / 2} } {2^{n / 2 - 1} \map \Gamma {\frac n 2} } \int_0^\infty v^{n / 2} e^{-v} \rd v\)
\(\ds \) \(=\) \(\ds 2 \times \frac {\map \Gamma {\frac n 2 + 1} } {\map \Gamma {\frac n 2} }\)
\(\ds \) \(=\) \(\ds 2 \times \frac {\frac n 2 \map \Gamma {\frac n 2} } {\map \Gamma {\frac n 2} }\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds n\)

So:

\(\ds \expect {\dfrac {Y / n} {Z / m} }\) \(=\) \(\ds \frac {n m} {n \paren {m - 2} }\)
\(\ds \) \(=\) \(\ds \frac m {m - 2}\)

$\blacksquare$